2nd Brazilian Mathematical Olympiad Problems 1980

2nd Brazilian Mathematical Olympiad Problems 1980

1.  Box A contains black balls and box B contains white balls. Take a certain number of balls from A and place them in B. Then take the same number of balls from B and place them in A. Is the number of white balls in A then greater, equal to, or less than the number of black balls in B?

2.  Show that for any positive integer n > 2 we can find n distinct positive integers such that the sum of their reciprocals is 1.

3.  Given a triangle ABC and a point P₀ on the side AB. Construct points P_i, Q_i, R_i as follows. Q_i is the foot of the perpendicular from P_i to BC, R_i is the foot of the perpendicular from Q_i to AC and P_i is the foot of the perpendicular from R_i-1 to AB. Show that the points P_i converge to a point P on AB and show how to construct P.

4.  Given 5 points of a sphere radius r, show that two of the points are a distance ≤ r √2 apart. 

Solutions

Problem 1

Box A contains black balls and box B contains white balls. Take a certain number of balls from A and place them in B. Then take the same number of balls from B and place them in A. Is the number of white balls in A then greater, equal to, or less than the number of black balls in B?
Answer
Equal.
Solution
Suppose we move n balls from A to B, then x white balls from B and n-x black balls from B. That leaves x black balls in B. So both numbers equal x.

Problem 2
Show that for any positive integer n > 2 we can find n distinct positive integers such that the sum of their reciprocals is 1.
Solution
We have 1/2 + 1/3 + 1/6 = 1. Now we have 1/2 + 1/4 + 1/8 + ... + 1/2ⁿ + 1/2ⁿ = 1, replace the second 1/2ⁿ by (1/2 + 1/3 + 1/6) 1/2ⁿ to get n+3 terms (n ≥ 1): 1/2 + 1/4 + 1/8 + ... + 1/2ⁿ⁺¹ + 1/(3·2ⁿ) + 1/(3·2ⁿ⁺¹) = 1. For example:
1/2 + 1/4 + 1/6 + 1/12 = 1
1/2 + 1/4 + 1/8 + 1/12 + 1/24 = 1

Problem 3
Given a triangle ABC and a point P₀ on the side AB. Construct points P_i, Q_i, R_i as follows. Q_i is the foot of the perpendicular from P_i to BC, R_i is the foot of the perpendicular from Q_i to AC and P_i is the foot of the perpendicular from R_i-1 to AB. Show that the points P_i converge to a point P on AB and show how to construct P.
Solution

It is clear from the diagram that Q_nQ_n+1 = P_nP_n+1 cos B, R_nR_n+1 = Q_nQ_n+1 cos C and P_nP_n+1 = R_n-1R_n cos A. Hence P_nP_n+1 = kⁿ P₀P₁, where |k| = |cos A cos B cos C| < 1. If we take the direction A to B as positive, then the signed distance P_nP_n+1 may be positive or negative, but the series 1 + |k| + |k|² + |k|³ + ... converges to 1/(1-|k|), so P₀P₁ + P₁P₂ + P₂P₃ + ... is absolutely convergent and hence convergent. So the points P_i converge to a point P on the line AB.

Take any point P' on AB, Take Q' as the foot of the perpendicular from P' to BC. Now take R' as the intersection of the lines through P' perpendicular to AB and through Q' perpendicular to AC. Now P'Q'R' is similar to the desired triangle PQR. Since B, P', P are collinear and B, Q', Q are collinear, it follows that B, R', R must be collinear. Thus extend BR' to meet AC at R. It is now straightforward to construct Q, then P. 

Problem 4
Given 5 points of a sphere radius r, show that two of the points are a distance ≤ r √2 apart.
Solution

Suppose the result is false so that we can find 5 points with the distance between any two > r√2. Then the angle subtended by any two at the center of the sphere is > 90^o. Take one of the points to be at the north pole. Then the other four must all be south of the equator. Two must have longitude differing by ≤ 90^o.

It is now fairly obvious that these two points subtend an angle ≤ 90^o at the center. To prove it we may take rectangular coordinates with origin at the center of the sphere so that both points have all coordinates non-negative. Suppose one is (a, b, c) and the other (A, B, C). Then since both lie on the sphere a² + b² + c² = A² + B² + C² = r², and the square of the distance between them is (a-A)² + (b-B)² + (c-C)² ≤ (a² + b² + c²) + (A² + B² + C²) = 2r², so the distance ≤ r√2, as required.